Animated Solution for Mathematics - Definite Integration: The value of ${\int }_{\pi /4}^{3\pi /4}\frac{\varphi }{1+\sin \varphi }d\varphi$ is $\dots \dots \dots$

Conquering the Definite Integral

Have you ever stared at an integral and felt like the variable in the numerator was mocking you? You are not alone.

Today, we are tackling a classic JEE Advanced problem: ${\int }_{\pi /4}^{3\pi /4}\frac{\varphi }{1+\sin \varphi }d\varphi$. At first glance, this looks like a nightmare.

The secret lies not in brute-force calculus, but in the elegance of symmetry.

The King's Property

In the world of JEE mathematics, there is a legendary tool known as the King's Property. It states that for any definite integral, ${\int }_a^{b}f(x)dx={\int }_a^{b}f(a+b-x)dx$.

This is powerful because it allows us to transform the integrand while keeping the area under the curve identical.

Look at our limits: $a=\frac{\pi }{4}$ and $b=\frac{3\pi }{4}$. Their sum is $a+b=\pi$.

If we replace $\varphi$ with $(\pi -\varphi )$, we are essentially looking at the function from the other side of the mirror. Let's define our integral as $I$:

Applying the property, we get:

Since $\sin (\pi -\varphi )=\sin \varphi$, our integral becomes:

The Algebraic Dance

Now, here is the magic. If we add our original integral to our transformed integral, we get $2I$. Because the denominators are identical, we simply add the numerators:

Watch closely—the $\varphi$ and $-\varphi$ cancel out perfectly! We are left with a constant $\pi$ in the numerator:

Suddenly, the variable that made the integral impossible to solve is gone. We have reduced a complex problem to a standard trigonometric integral.

The Trigonometric Transformation

How do we handle $\frac{1}{1+\sin \varphi }$? We use the conjugate trick by multiplying the numerator and denominator by $(1-\sin \varphi )$:

Using the identity $1-{\sin }^2\varphi ={\cos }^2\varphi$, we get:

Now, split the fraction into two parts:

These are standard integrals. The integral of ${\sec }^2\varphi$ is $\tan \varphi$, and the integral of $\sec \varphi \tan \varphi$ is $\sec \varphi$.

The Final Evaluation

We are in the home stretch. We evaluate the expression $[\tan \varphi -\sec \varphi ]$ from $\frac{\pi }{4}$ to $\frac{3\pi }{4}$:

Plugging in the values: $\tan \frac{3\pi }{4}=-1$, $\sec \frac{3\pi }{4}=-\sqrt{2}$, $\tan \frac{\pi }{4}=1$, and $\sec \frac{\pi }{4}=\sqrt{2}$.

Dividing by $2$, we arrive at our destination:

Isn't that beautiful? We started with a daunting expression and, through the symmetry of the King's Property, stripped away the complexity layer by layer. Keep this technique in your toolkit—whenever you see a linear variable in the numerator of a definite integral, think of the King!